lecture 10 (calculus of polar curves)
TRANSCRIPT
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
1/184
Tangent Lines to Polar Curves
Calculus of Polar Curves
Institute of Mathematics, University of the Philippines Diliman
Mathematics 54 (Elementary Analysis 2)
Polar Curves 1/ 18
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
2/184
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
Polar Curves 2/ 18
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
3/184
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
Polar Curves 2/ 18
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
4/184
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
Polar Curves 2/ 18
l
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
5/184
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
Polar Curves 2/ 18
T t Li t P l C
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
6/184
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
Polar Curves 2/ 18
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
7/184
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
Goal: obtain slopes of tangent lines to polar curves of form r=f()
Polar Curves 2/ 18
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
8/184
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
Parametrization of a Polar Curve
A polar curve r=f() can be parametrized as
Polar Curves 3/ 18
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
9/184
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
Parametrization of a Polar Curve
A polar curve r=f() can be parametrized asx = rcos
Polar Curves 3/ 18
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
10/184
a ge t es to o a Cu es
Tangent Lines to Polar Curves
Parametrization of a Polar Curve
A polar curve r=f() can be parametrized asx = rcos = f()cos
Polar Curves 3/ 18
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
11/184
g
Tangent Lines to Polar Curves
Parametrization of a Polar Curve
A polar curve r=f() can be parametrized asx = rcos = f()cosy = rsin
Polar Curves 3/ 18
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
12/184
Tangent Lines to Polar Curves
Parametrization of a Polar Curve
A polar curve r=f() can be parametrized asx = rcos = f()cosy = rsin = f()sin
Polar Curves 3/ 18
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
13/184
Tangent Lines to Polar Curves
Parametrization of a Polar Curve
A polar curve r=f() can be parametrized asx = rcos = f()cosy = rsin = f()sin
Recall. slope of a parametric curvedydx
=
dyddx
d
Polar Curves 3/ 18
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
14/184
Tangent Lines to Polar Curves
Parametrization of a Polar Curve
A polar curve r=f() can be parametrized asx = rcos = f()cosy = rsin = f()sin
Recall. slope of a parametric curvedydx
=
dyddx
d
dx
d=f() cosf()sin,
Polar Curves 3/ 18
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
15/184
Tangent Lines to Polar Curves
Parametrization of a Polar Curve
A polar curve r=f() can be parametrized asx = rcos = f()cosy = rsin = f()sin
Recall. slope of a parametric curvedydx
=
dyddx
d
dx
d=f() cosf()sin, dy
d=f() sin+f()cos
Polar Curves 3/ 18
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
16/184
Tangent Lines to Polar Curves
Parametrization of a Polar Curve
A polar curve r=f() can be parametrized asx = rcos = f()cosy = rsin = f()sin
Recall. slope of a parametric curvedydx
=
dyddx
d
dx
d=f() cosf()sin, dy
d=f() sin+f()cos
Slope of a Tangent Line to a Polar Curve
Given that dy/d and dx/d are continuous and dx/d = 0, then the slope of thepolar curve r=f()is
dy
dx=
drd
sin+ rcosdrd
cos rsin
Polar Curves 3/ 18
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
17/184
Tangent Lines to Polar Curves
Example.
Find the (Cartesian) equation of the tangent line to the cardioid r
=1
+sin at the
point where = 3 .
Polar Curves 4/ 18
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
18/184
Tangent Lines to Polar Curves
Example.
Find the (Cartesian) equation of the tangent line to the cardioid r
=1
+sin at the
point where = 3 .
Solution. Recall thatdydx
=drd
sin+rcosdrd
cosrsin
Polar Curves 4/ 18
Tangent Lines to Polar Curves
l
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
19/184
Tangent Lines to Polar Curves
Example.
Find the (Cartesian) equation of the tangent line to the cardioid r
=1
+sin at the
point where = 3 .
Solution. Recall thatdydx
=drd
sin+rcosdrd
cosrsin
Meanwhile, we have the following:
r= 1 +sin 3
Polar Curves 4/ 18
Tangent Lines to Polar Curves
T Li P l C
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
20/184
Tangent Lines to Polar Curves
Example.
Find the (Cartesian) equation of the tangent line to the cardioid r
=1
+sin at the
point where = 3 .
Solution. Recall thatdydx
=drd
sin+rcosdrd
cosrsin
Meanwhile, we have the following:
r= 1 +sin 3
= 1+ 32
Polar Curves 4/ 18
Tangent Lines to Polar Curves
T t Li t P l C
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
21/184
Tangent Lines to Polar Curves
Example.
Find the (Cartesian) equation of the tangent line to the cardioid r
=1
+sin at the
point where = 3 .
Solution. Recall thatdydx
=drd
sin+rcosdrd
cosrsin
Meanwhile, we have the following:
r= 1 +sin 3
= 1+ 32 = 2+32
Polar Curves 4/ 18
Tangent Lines to Polar Curves
T t Li t P l C
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
22/184
Tangent Lines to Polar Curves
Example.
Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where = 3 .
Solution. Recall thatdydx
=drd
sin+rcosdrd
cosrsin
Meanwhile, we have the following:
r= 1 +sin 3
= 1+ 32 = 2+32drd
Polar Curves 4/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
23/184
Tangent Lines to Polar Curves
Example.
Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where = 3 .
Solution. Recall thatdydx
=drd
sin+rcosdrd
cosrsin
Meanwhile, we have the following:
r= 1 +sin 3
= 1+ 32 = 2+32drd
= cos
3
Polar Curves 4/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
24/184
Tangent Lines to Polar Curves
Example.
Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where = 3 .
Solution. Recall thatdydx
=drd
sin+rcosdrd
cosrsin
Meanwhile, we have the following:
r= 1 +sin 3
= 1+ 32 = 2+32drd
= cos
3
= 12
Polar Curves 4/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
25/184
Tangent Lines to Polar Curves
Example.
Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where = 3 .
Solution. Recall thatdydx
=drd
sin+rcosdrd
cosrsin
Meanwhile, we have the following:
r= 1 +sin 3
= 1+ 32 = 2+32drd
= cos
3
= 12
cos3
Polar Curves 4/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
26/184
Tangent Lines to Polar Curves
Example.
Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where = 3 .
Solution. Recall thatdydx
=drd
sin+rcosdrd
cosrsin
Meanwhile, we have the following:
r= 1 +sin 3
= 1+ 32 = 2+32drd
= cos
3
= 12
cos3
= 12
Polar Curves 4/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
27/184
Tangent Lines to Polar Curves
Example.
Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where = 3 .
Solution. Recall thatdydx
=drd
sin+rcosdrd
cosrsin
Meanwhile, we have the following:
r= 1 +sin 3
= 1+ 32 = 2+32drd
= cos
3
= 12
cos3
= 12
sin
3=
Polar Curves 4/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
28/184
Tangent Lines to Polar Curves
Example.
Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where = 3 .
Solution. Recall thatdydx
=drd
sin+rcosdrd
cosrsin
Meanwhile, we have the following:
r= 1 +sin 3
= 1+ 32 = 2+32drd
= cos
3
= 12
cos3
= 12
sin
3=
32
Polar Curves 4/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
29/184
Tangent Lines to Polar Curves
Example.
Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where =
3 .
Solution. Recall thatdydx
=drd
sin+rcosdrd
cosrsin
Meanwhile, we have the following:
r= 1 +sin 3
= 1+ 32 = 2+32drd
= cos
3
= 12
cos3
= 12
sin
3=
32
Hence,
dy
dx=
Polar Curves 4/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
30/184
g C
Example.
Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where =
3 .
Solution. Recall thatdydx
=drd
sin+rcosdrd
cosrsin
Meanwhile, we have the following:
r= 1 +sin 3
= 1+ 32 = 2+32drd
= cos
3
= 12
cos3
= 12
sin
3=
32
Hence,
dy
dx=
12
Polar Curves 4/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
31/184
g
Example.
Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where =
3 .
Solution. Recall thatdydx
=drd
sin+rcosdrd
cosrsin
Meanwhile, we have the following:
r= 1 +sin 3
= 1+ 32 = 2+32drd
= cos
3
= 12
cos3
= 12
sin
3=
32
Hence,
dy
dx=
12
3
2
+
Polar Curves 4/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
32/184
g
Example.
Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where =
3 .
Solution. Recall thatdydx
=drd
sin+rcosdrd
cosrsin
Meanwhile, we have the following:
r= 1 +sin 3
= 1+ 32 = 2+32drd
= cos
3
= 12
cos3
= 12
sin
3=
3
2
Hence,
dy
dx=
12
3
2
+ 2+
3
2
Polar Curves 4/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
33/184
g
Example.
Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where =
3 .
Solution. Recall thatdydx
=drd
sin+rcosdrd
cosrsin
Meanwhile, we have the following:
r= 1 +sin 3
= 1+ 32 = 2+32drd
= cos
3
= 12
cos3
= 12
sin
3=
3
2
Hence,
dy
dx=
12
3
2
+ 2+
3
2
12
Polar Curves 4/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
34/184
Example.
Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where =
3 .
Solution. Recall thatdydx
=drd
sin+rcosdrd
cosrsin
Meanwhile, we have the following:
r= 1 +sin 3
= 1+ 32 = 2+32drd
= cos
3
= 12
cos3
= 12
sin
3=
3
2
Hence,
dy
dx=
12
3
2
+ 2+
3
2
12
12
Polar Curves 4/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
35/184
Example.
Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where =
3 .
Solution. Recall thatdydx
=drd
sin+rcosdrd
cosrsin
Meanwhile, we have the following:
r= 1 +sin 3
= 1+ 32 = 2+32drd
= cos
3
= 12
cos3
= 12
sin3
=
32
Hence,
dy
dx=
12
3
2
+ 2+
3
2
12
12
12
Polar Curves 4/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
36/184
Example.
Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where =
3 .
Solution. Recall thatdydx
=drd
sin+rcosdrd
cosrsin
Meanwhile, we have the following:
r= 1 +sin 3
= 1+ 32 = 2+32drd
= cos
3
= 12
cos3
= 12
sin3
=
32
Hence,
dy
dx=
12
3
2
+ 2+
3
2
12
12
12
2+
32
Polar Curves 4/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
37/184
Example.
Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where =
3 .
Solution. Recall thatdydx
=drd
sin+rcosdrd
cosrsin
Meanwhile, we have the following:
r= 1 +sin 3
= 1+ 32 = 2+32drd
= cos
3
= 12
cos3
= 12
sin3
=
32
Hence,
dy
dx=
12
3
2
+ 2+
3
2
12
12
12
2+
32
3
2 Polar Curves 4/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
38/184
Example.
Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where =
3 .
Solution. Recall thatdydx
=drd
sin+rcosdrd
cosrsin
Meanwhile, we have the following:
r= 1 +sin 3
= 1+ 32 = 2+32drd
= cos
3
= 12
cos3
= 12
sin3
=
32
Hence,
dy
dx=
12
3
2
+ 2+
3
2
12
12
12
2+
3
2
3
2
= 1+
32
1
32
Polar Curves 4/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
39/184
Example.
Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where =
3 .
Solution. Recall thatdydx
=drd
sin+rcosdrd
cosrsin
Meanwhile, we have the following:
r= 1 +sin 3
= 1+
32 = 2+
32
drd
= cos
3
= 12
cos3
= 12
sin3
=
32
Hence,
dy
dx=
12
3
2
+ 2+
3
2
12
12
12
2+
3
2
3
2
= 1+
32
1
32
= 1
Polar Curves 4/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
40/184
Example.
Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at thepoint where =
3 .
Solution. Recall thatdydx
=drd
sin+rcosdrd
cosrsin
Meanwhile, we have the following:
r= 1 +sin 3
= 1+
32 = 2+
32
drd
= cos
3
= 12
cos3
= 12
sin3
=
32
Hence,
dy
dx=
12
3
2
+ 2+
3
2
12
12
12
2+
3
2
3
2
= 1+
32
1
32
= 1
The point of tangency
2+
32 ,
3
in
Cartesian is
Polar Curves 4/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
41/184
Example.
Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at the
point where =
3 .
Solution. Recall thatdydx
=drd
sin+rcosdrd
cosrsin
Meanwhile, we have the following:
r= 1 +sin 3
= 1+
32 = 2+
32
drd
= cos
3
= 12
cos3
= 12
sin3 =
3
2
Hence,
dy
dx=
12
3
2
+ 2+
3
2
12
12
12
2+
3
2
3
2
= 1+
32
1
32
= 1
The point of tangency
2+
32 ,
3
in
Cartesian is
2+
34 ,
3+2
34
.
Polar Curves 4/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
42/184
Example.
Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at the
point where =
3 .
Solution. Recall thatdydx
=drd
sin+rcosdrd
cosrsin
Meanwhile, we have the following:
r= 1 +sin 3
= 1+
32 = 2+
32
drd
= cos
3
= 12
cos3
= 12
sin3 =
3
2
Hence,
dy
dx=
12
3
2
+ 2+
3
2
12
12
12
2+
3
2
3
2
= 1+
32
1
32
= 1
The point of tangency
2+
32 ,
3
in
Cartesian is
2+
34 ,
3+2
34
.
Thus, the equation is
y
3+2
3
4= x
2+
34
Polar Curves 4/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
43/184
Example.
Find the (Cartesian) equation of the tangent line to the cardioid r= 1 +sin at the
point where =
3 .
Solution. Recall thatdydx
=drd
sin+rcosdrd
cosrsin
Meanwhile, we have the following:
r= 1 +sin 3
= 1+
32 = 2+
32
drd
= cos
3
= 12
cos3
= 12
sin3 =
3
2
Hence,
dy
dx=
12
3
2
+ 2+
3
2
12
12
12
2+
3
2
3
2
= 1+
32
1
32
= 1
The point of tangency
2+
32 ,
3
in
Cartesian is
2+
34 ,
3+2
34
.
Thus, the equation is
y
3+2
3
4= x
2+
34
Polar Curves 4/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
44/184
Example.
Determine the points on the cardioid r= 1 coswhere the tangent line ishorizontal, or vertical.
Polar Curves 5/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
45/184
Example.
Determine the points on the cardioid r= 1 coswhere the tangent line ishorizontal, or vertical.
Recall. For a parametric curve
C : x
=x(), y
=y(),
dyd
= 0
Polar Curves 5/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
46/184
Example.
Determine the points on the cardioid r= 1 coswhere the tangent line ishorizontal, or vertical.
Recall. For a parametric curve
C : x
=x(), y
=y(),
dyd
= 0 = horizontal tangent line
Polar Curves 5/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
47/184
Example.
Determine the points on the cardioid r= 1 coswhere the tangent line ishorizontal, or vertical.
Recall. For a parametric curve
C : x
=x(), y
=y(),
dyd
= 0 = horizontal tangent linedxd
= 0
Polar Curves 5/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
48/184
Example.
Determine the points on the cardioid r= 1 coswhere the tangent line ishorizontal, or vertical.
Recall. For a parametric curve
C : x
=x(), y
=y(),
dyd
= 0 = horizontal tangent linedxd
= 0 = vertical tangent line
Polar Curves 5/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
49/184
Example.
Determine the points on the cardioid r= 1 coswhere the tangent line ishorizontal, or vertical.
Recall. For a parametric curve
C : x
=x(), y
=y(),
dyd
= 0 = horizontal tangent linedxd
= 0 = vertical tangent lineprovided that they are not simultaneously zero.
Polar Curves 5/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
50/184
Example.
Determine the points on the cardioid r= 1 coswhere the tangent line ishorizontal, or vertical.
Recall. For a parametric curve
C : x
=x(), y
=y(),
dyd
= 0 = horizontal tangent linedxd
= 0 = vertical tangent lineprovided that they are not simultaneously zero.
Note that for the above curve, we have the following parametrization:
Polar Curves 5/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
51/184
Example.
Determine the points on the cardioid r= 1 coswhere the tangent line ishorizontal, or vertical.
Recall. For a parametric curve
C : x
=x(), y
=y(),
dyd
= 0 = horizontal tangent linedxd
= 0 = vertical tangent lineprovided that they are not simultaneously zero.
Note that for the above curve, we have the following parametrization:
x = rcos
Polar Curves 5/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
52/184
Example.
Determine the points on the cardioid r= 1 coswhere the tangent line ishorizontal, or vertical.
Recall. For a parametric curve
C : x
=x(), y
=y(),
dyd
= 0 = horizontal tangent linedxd
= 0 = vertical tangent lineprovided that they are not simultaneously zero.
Note that for the above curve, we have the following parametrization:
x = rcos = (1 cos)cos
Polar Curves 5/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
53/184
Example.
Determine the points on the cardioid r= 1 coswhere the tangent line ishorizontal, or vertical.
Recall. For a parametric curve
C : x
=x(), y
=y(),
dyd
= 0 = horizontal tangent linedxd
= 0 = vertical tangent lineprovided that they are not simultaneously zero.
Note that for the above curve, we have the following parametrization:
x = rcos = (1 cos)cos= coscos2
Polar Curves 5/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
54/184
Example.
Determine the points on the cardioid r= 1 coswhere the tangent line ishorizontal, or vertical.
Recall. For a parametric curve
C : x
=x(), y
=y(),
dyd
= 0 = horizontal tangent linedxd
= 0 = vertical tangent lineprovided that they are not simultaneously zero.
Note that for the above curve, we have the following parametrization:
x = rcos = (1 cos)cos= coscos2y = rsin
Polar Curves 5/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
55/184
Example.
Determine the points on the cardioid r= 1 coswhere the tangent line ishorizontal, or vertical.
Recall. For a parametric curve
C : x= x(), y=y(),dyd
= 0 = horizontal tangent linedxd
= 0 = vertical tangent lineprovided that they are not simultaneously zero.
Note that for the above curve, we have the following parametrization:
x = rcos = (1 cos)cos= coscos2y = rsin= (1 cos)sin
Polar Curves 5/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
56/184
Example.
Determine the points on the cardioid r= 1 coswhere the tangent line ishorizontal, or vertical.
Recall. For a parametric curve
C : x= x(), y=y(),dyd
= 0 = horizontal tangent linedxd
= 0 = vertical tangent lineprovided that they are not simultaneously zero.
Note that for the above curve, we have the following parametrization:
x = rcos = (1 cos)cos= coscos2y = rsin= (1 cos)sin= sinsincos
Polar Curves 5/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
E l ti d
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
57/184
Example. continued...
Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.
dx
d=
Polar Curves 6/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
Example continued
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
58/184
Example. continued...
Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.
dx
d= sin+2cos sin
Polar Curves 6/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
Example continued
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
59/184
Example. continued...
Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.
dx
d= sin+2cos sin= sin(2cos1)
Polar Curves 6/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
Example continued
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
60/184
Example. continued...
Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.
dx
d= sin+2cos sin= sin(2cos1)
=0
Polar Curves 6/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
Example continued
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
61/184
Example. continued...
Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.
dx
d= sin+2cos sin= sin(2cos1)
=0
=sin
=0 or cos
=1
2
Polar Curves 6/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
Example continued
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
62/184
Example. continued...
Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.
dx
d= sin+2cos sin= sin(2cos1)
=0
=sin
=0 or cos
=1
2 =
=0,
Polar Curves 6/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
Example. continued...
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
63/184
Example. continued...
Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.
dx
d= sin+2cos sin= sin(2cos1)
=0
=sin
=0 or cos
=1
2 =
=0,,
Polar Curves 6/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
Example. continued...
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
64/184
p
Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.
dx
d= sin+2cos sin= sin(2cos1)
=0
=sin
=0 or cos
=1
2 =
=0,,
3
,
Polar Curves 6/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
Example. continued...
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
65/184
p
Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.
dx
d= sin+2cos sin= sin(2cos1)
=0
=sin
=0 or cos
=1
2 =
=0,,
3
,5
3
Polar Curves 6/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
Example. continued...
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
66/184
p
Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.
dx
d= sin+2cos sin= sin(2cos1)
=0
=sin
=0 or cos
=1
2 =
=0,,
3
,5
3
dy
d=
Polar Curves 6/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
Example. continued...
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
67/184
Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.
dx
d= sin+2cos sin= sin(2cos1)
=0
=sin
=0 or cos
=1
2 =
=0,,
3
,5
3
dy
d= cos (cos2 sin2)
Polar Curves 6/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
Example. continued...
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
68/184
Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.
dx
d= sin+2cos sin= sin(2cos1)
=0
=sin
=0 or cos
=
1
2 =
=0,,
3
,5
3
dy
d= cos (cos2 sin2)
= cos 2cos2+1
Polar Curves 6/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
Example. continued...
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
69/184
Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.
dx
d= sin+2cos sin= sin(2cos1)
=0
=sin
=0 or cos
=
1
2 =
=0,,
3
,5
3
dy
d= cos (cos2 sin2)
= cos 2cos2+1
=(1
cos)(1
+2cos)
Polar Curves 6/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
Example. continued...
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
70/184
Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.
dx
d= sin+2cos sin= sin(2cos1)
=0
=sin
=0 or cos
=
1
2 =
=0,,
3
,5
3
dy
d= cos (cos2 sin2)
= cos 2cos2+1
=(1
cos)(1
+2cos)
= 0
Polar Curves 6/ 18
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
71/184
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
Example. continued...
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
72/184
Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.
dx
d= sin+2cos sin= sin(2cos1)
=0
=sin
=0 or cos
=
1
2 =
=0,,
3
,5
3
dy
d= cos (cos2 sin2)
= cos 2cos2+1
=(1
cos)(1
+2cos)
= 0 = cos= 1 or cos = 12
= = 0,
Polar Curves 6/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
Example. continued...
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
73/184
Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.
dx
d= sin+2cos sin= sin(2cos1)
=0
=sin
=0 or cos
=
1
2 =
=0,,
3
,5
3
dy
d= cos (cos2 sin2)
= cos 2cos2+1
=(1
cos)(1
+2cos)
= 0 = cos= 1 or cos = 12
= = 0, 23
,
Polar Curves 6/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
Example. continued...
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
74/184
Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.
dx
d= sin+2cos sin= sin(2cos1)
=0
=sin
=0 or cos
=
1
2 =
=0,,
3
,5
3
dy
d= cos (cos2 sin2)
= cos 2cos2+1
=(1
cos)(1
+2cos)
= 0 = cos= 1 or cos = 12
= = 0, 23
, 43
Polar Curves 6/ 18
Tangent Lines to Polar Curves
Tangent Lines to Polar Curves
Example. continued...
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
75/184
Determine the points on the cardioid C : x= coscos2, y= sin sincoswhere the tangent line is horizontal, or vertical.
dx
d= sin+2cos sin= sin(2cos1)
=0
=sin
=0 or cos
=
1
2 =
=0,,
3
,5
3
dy
d= cos (cos2 sin2)
= cos 2cos2+1
=(1
cos)(1
+2cos)
= 0 = cos= 1 or cos = 12
= = 0, 23
, 43
Thus, tangent lines are horizontal at
32 ,
23
and
32 ,
43
, and are vertical at
12 ,
3
,
12 ,
53 and (2,).
Polar Curves 6/ 18
Tangent Lines to Polar Curves
Arc Length for Polar Curves
Recall. arc length for a parametric curve
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
76/184
Polar Curves 7/ 18
Tangent Lines to Polar Curves
Arc Length for Polar Curves
Recall. arc length for a parametric curve
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
77/184
L=b
adx
dt2 +
dy
dt2
dt.
Polar Curves 7/ 18
Tangent Lines to Polar Curves
Arc Length for Polar Curves
Recall. arc length for a parametric curve
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
78/184
L=b
adx
dt2 +
dy
dt2
dt.
For a polar curve r=f(), parametrized as x=f()cos, y=f()sin,
dx
d 2
Polar Curves 7/ 18
Tangent Lines to Polar Curves
Arc Length for Polar Curves
Recall. arc length for a parametric curve
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
79/184
L=
b
adx
dt2 +
dy
dt2 dt.
For a polar curve r=f(), parametrized as x=f()cos, y=f()sin,
dx
d 2
= dr
dcos rsin
2
Polar Curves 7/ 18
Tangent Lines to Polar Curves
Arc Length for Polar Curves
Recall. arc length for a parametric curve
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
80/184
L=
b
adx
dt2 +
dy
dt2 dt.
For a polar curve r=f(), parametrized as x=f()cos, y=f()sin,
dx
d 2
= dr
dcos rsin
2
=
dr
d
2cos22rdr
dcos sin+ r2 sin2
Polar Curves 7/ 18
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
81/184
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
82/184
Tangent Lines to Polar Curves
Arc Length for Polar Curves
Recall. arc length for a parametric curve
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
83/184
L=
b
adx
dt2 +
dy
dt2 dt.
For a polar curve r=f(), parametrized as x=f()cos, y=f()sin,
dx
d 2
= dr
dcos rsin
2
=
dr
d
2cos22rdr
dcos sin+ r2 sin2
dy
d
2=
dr
dsin+ rcos
2
= drd
2sin2+2r
dr
d cos sin+ r2 cos2
Polar Curves 7/ 18
Tangent Lines to Polar Curves
Arc Length for Polar Curves
Recall. arc length for a parametric curve
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
84/184
L=
b
adx
dt
2
+dy
dt
2
dt.
For a polar curve r=f(), parametrized as x=f()cos, y=f()sin,
dx
d 2
= dr
dcos rsin
2
=
dr
d
2cos22rdr
dcos sin+ r2 sin2
dy
d
2=
dr
dsin+ rcos
2
= drd
2sin2+2r
dr
d cos sin+ r2 cos2
Thus,
dxd
2 + dyd
2=
Polar Curves 7/ 18
Tangent Lines to Polar Curves
Arc Length for Polar Curves
Recall. arc length for a parametric curve
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
85/184
L=
b
adx
dt
2
+dy
dt
2
dt.
For a polar curve r=f(), parametrized as x=f()cos, y=f()sin,
dx
d 2
= dr
dcos rsin
2
=
dr
d
2cos22rdr
dcos sin+ r2 sin2
dy
d
2=
dr
dsin+ rcos
2
= drd 2
sin2+2rdr
d cos sin+ r2 cos2
Thus,
dxd
2 + dyd
2= r2 +
drd
2
Polar Curves 7/ 18
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
86/184
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
87/184
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
88/184
Tangent Lines to Polar Curves
Arc Length for Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
89/184
Example.
Find the length of the circle r= 4cos.
As varies from 0 to 2
Polar Curves 9/ 18
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
90/184
Tangent Lines to Polar Curves
Arc Length for Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
91/184
Example.
Find the length of the circle r= 4cos.
As varies from 2 to
Polar Curves 9/ 18 Tangent Lines to Polar Curves
Arc Length for Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
92/184
Example.
Find the length of the circle r= 4cos.
must vary from 0 to L=
0 (4cos)2 + (4sin)2 d
Polar Curves 9/ 18 Tangent Lines to Polar Curves
Arc Length for Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
93/184
Example.
Find the length of the circle r= 4cos.
must vary from 0 to L=
0 (4cos)2 + (4sin)2 d
=
0
16cos2+ 16sin2d
Polar Curves 9/ 18
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
94/184
Tangent Lines to Polar Curves
Arc Length for Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
95/184
Example.
Find the length of the circle r= 4cos.
must vary from 0 to L=
0 (4cos)2 + (4sin)2 d
=
0
16cos2+ 16sin2d
=
04 d
= 4
Polar Curves 9/ 18 Tangent Lines to Polar Curves
Arc Length for Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
96/184
Example.Setup the integral that yields the length of one petal of the rose r= sin 2.
Polar Curves 10/ 18 Tangent Lines to Polar Curves
Arc Length for Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
97/184
Example.Setup the integral that yields the length of one petal of the rose r= sin 2.
As varies from 0 to 4
Polar Curves 10/ 18
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
98/184
Tangent Lines to Polar Curves
Arc Length for Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
99/184
Example.
Setup the integral that yields the length of one petal of the rose r= sin 2.
As varies from 4 to2
Polar Curves 10/ 18 Tangent Lines to Polar Curves
Arc Length for Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
100/184
Example.
Setup the integral that yields the length of one petal of the rose r= sin 2.
As varies from 4 to2
Polar Curves 10/ 18
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
101/184
Tangent Lines to Polar Curves
Arc Length for Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
102/184
Example.
Setup the integral that yields the length of one petal of the rose r= sin 2.
must vary from 0 to 2L=
/20
(sin 2)2 + (2cos 2)2 d
=/20
sin2 2+4cos2 2d
Polar Curves 10/ 18
Tangent Lines to Polar Curves
Arc Length for Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
103/184
Example.
Setup the integral that yields the length of one petal of the rose r= sin 2.
must vary from 0 to 2L=
/20
(sin 2)2 + (2cos 2)2 d
=/20
sin2 2+4cos2 2d
=/2
0
1 +3cos2 2d
Polar Curves 10/ 18
Tangent Lines to Polar Curves
Arc Length for Polar Curves
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
104/184
Example.
Setup the integral that yields the length of one petal of the rose r= sin 2.
must vary from 0 to 2L=
/20
(sin 2)2 + (2cos 2)2 d
=/20
sin2 2+4cos2 2d
=/2
0
1 +3cos2 2d
or, by symmetry,
=2
/4
01 + 3cos
2 2d
Polar Curves 10/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Consider the following polar region.
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
105/184
Polar Curves 11/ 18
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
106/184
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
107/184
Tangent Lines to Polar Curves
Area in Polar Coordinates
Consider the following polar region.
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
108/184
Assume that + 2.
Polar Curves 11/ 18
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
109/184
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
110/184
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
111/184
Tangent Lines to Polar Curves
Area in Polar Coordinates
Consider the following polar region.
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
112/184
Assume that + 2.Divide R into n wedgesusing the rays
=1,
=2, .. .,
=n
1.
Let A1, A2, .. ., An be the areas of the wedges.
Polar Curves 11/ 18
Tangent Lines to Polar Curves
Area in Polar CoordinatesConsider the following polar region.
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
113/184
Assume that + 2.Divide R into n wedgesusing the rays
=1,
=2, .. .,
=n
1.
Let A1, A2, .. ., An be the areas of the wedges.
Let 1, 2, .. ., n be the central angles of the wedges.
Polar Curves 11/ 18
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
114/184
Tangent Lines to Polar Curves
Area in Polar CoordinatesConsider the following polar region.
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
115/184
Assume that + 2.Divide R into n wedgesusing the rays
=1,
=2, .. .,
=n
1.
Let A1, A2, .. ., An be the areas of the wedges.
Let 1, 2, .. ., n be the central angles of the wedges.
The area Aof the region is given byA=A1 +A2 + +An=n
i=1Ai.
Polar Curves 11/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Approximate the area of the ith wedge by that of a sector with central angle i,
and radius f(i
), where i
[i1,i].
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
116/184
Polar Curves 12/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Approximate the area of the ith wedge by that of a sector with central angle i,
and radius f(i
), where i
[i1,i].
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
117/184
Polar Curves 12/ 18
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
118/184
Tangent Lines to Polar Curves
Area in Polar Coordinates
Approximate the area of the ith wedge by that of a sector with central angle i,
and radius f(i
), where i
[i1,i].
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
119/184
Thus, Ai
Polar Curves 12/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Approximate the area of the ith wedge by that of a sector with central angle i,
and radius f(i
), where i
[i1,i].
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
120/184
Thus, Ai1
2f(i )
2i.
Polar Curves 12/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Approximate the area of the ith wedge by that of a sector with central angle i,
and radius f(i
), where i
[i1,i].
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
121/184
Thus, Ai1
2f(i )
2i. And A=
ni=1Ai
ni=1
1
2f(i )2i.
Polar Curves 12/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Approximate the area of the ith wedge by that of a sector with central angle i,
and radius f(i
), where i
[i1,i].
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
122/184
Thus, Ai1
2f(i )
2i. And A=
ni=1Ai
ni=1
1
2f(i )2i.
If we increase nin such a way that i 0, for everyi= 1,2,..., n, then
Polar Curves 12/ 18
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
123/184
Tangent Lines to Polar Curves
Area in Polar Coordinates
Approximate the area of the ith wedge by that of a sector with central angle i,
and radius f(i
), where i
[i1,i].
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
124/184
Thus, Ai1
2f(i )
2
i. And A=n
i=1Ain
i=1
1
2f(i )2i.
If we increase nin such a way that i 0, for everyi= 1,2,..., n, thenA= lim
nn
i=1
1
2
f(i )
2i=
1
2
f()
2d.
Polar Curves 12/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Area of a Polar Region.
Let and be angles such that +2. Let r=f() be continuous on [,].The area of the region
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
125/184
is given by
A=
1
2[f()]2d =
1
2r2d.
Polar Curves 13/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Determine the area of the region in the first quadrant inside the cardioid
r= 1 cos.
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
126/184
Polar Curves 14/ 18
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
127/184
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
128/184
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
129/184
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Determine the area of the region in the first quadrant inside the cardioid
r= 1 cos.
Solution The region is
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
130/184
Solution. The region is
The limits of integration are determined by the radial lines, or rays, that bound the
region.
Polar Curves 14/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Determine the area of the region in the first quadrant inside the cardioid
r= 1 cos.
Solution The region is
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
131/184
Solution. The region is
The limits of integration are determined by the radial lines, or rays, that bound the
region. Here, must vary from = 0
Polar Curves 14/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Determine the area of the region in the first quadrant inside the cardioid
r= 1 cos.
Solution The region is
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
132/184
Solution. The region is
The limits of integration are determined by the radial lines, or rays, that bound the
region. Here, must vary from = 0 to =/2 to sweep out the region.
Polar Curves 14/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Determine the area of the region in the first quadrant inside the cardioid
r= 1 cos.
Solution. The region is
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
133/184
Solution. The region is
The limits of integration are determined by the radial lines, or rays, that bound the
region. Here, must vary from = 0 to =/2 to sweep out the region. Thus
A =/2
012
(1 cos)2 d
Polar Curves 14/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Determine the area of the region in the first quadrant inside the cardioid
r= 1 cos.
Solution. The region is
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
134/184
Solution. The region is
The limits of integration are determined by the radial lines, or rays, that bound the
region. Here, must vary from = 0 to =/2 to sweep out the region. Thus
A = /20
12
(1 cos)2 d=/20
12
(1 2cos+cos2) d
Polar Curves 14/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Determine the area of the region in the first quadrant inside the cardioid
r= 1 cos.
Solution. The region is
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
135/184
g
The limits of integration are determined by the radial lines, or rays, that bound the
region. Here, must vary from = 0 to =/2 to sweep out the region. Thus
A = /20
12
(1 cos)2 d=/20
12
(1 2cos+cos2) d
= 12
/20
3
2 2cos+ 1
2cos2
d
Polar Curves 14/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Determine the area of the region in the first quadrant inside the cardioid
r= 1 cos.
Solution. The region is
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
136/184
g
The limits of integration are determined by the radial lines, or rays, that bound the
region. Here, must vary from = 0 to =/2 to sweep out the region. Thus
A = /20
12
(1 cos)2 d=/20
12
(1 2cos+cos2) d
= 12
/20
3
2 2cos+ 1
2cos2
d= 1
2
3
22sin+ 1
4sin2
/20
Polar Curves 14/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Determine the area of the region in the first quadrant inside the cardioid
r= 1 cos.
Solution. The region is
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
137/184
g
The limits of integration are determined by the radial lines, or rays, that bound the
region. Here, must vary from = 0 to =/2 to sweep out the region. Thus
A = /20
12
(1 cos)2 d=/20
12
(1 2cos+cos2) d
= 12
/20
3
2 2cos+ 1
2cos2
d= 1
2
3
22sin+ 1
4sin2
/20
= 38
1
Polar Curves 14/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region enclosed by the rose curve r= 3cos2.
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
138/184
Polar Curves 15/ 18
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
139/184
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
140/184
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region enclosed by the rose curve r= 3cos2.
Solution. The region is
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
141/184
Polar Curves 15/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region enclosed by the rose curve r= 3cos2.
Solution. The region is
We use s mmetr
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
142/184
We use symmetry.
Polar Curves 15/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region enclosed by the rose curve r= 3cos2.
Solution. The region is
We use symmetry
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
143/184
We use symmetry.
Here, must vary from = 0
Polar Curves 15/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region enclosed by the rose curve r= 3cos2.
Solution. The region is
We use symmetry
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
144/184
We use symmetry.
Here, must vary from = 0 to =/4to sweep out the region.
Polar Curves 15/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region enclosed by the rose curve r= 3cos2.
Solution. The region is
We use symmetry
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
145/184
We use symmetry.
Here, must vary from = 0 to =/4to sweep out the region. Thus
A = 8/4
0
1
2(3cos2)2 d
Polar Curves 15/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region enclosed by the rose curve r= 3cos2.
Solution. The region is
We use symmetry
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
146/184
We use symmetry.
Here, must vary from = 0 to =/4to sweep out the region. Thus
A = 8/4
0
1
2(3cos2)2 d= 36
/40
1
2(1 +cos4) d
Polar Curves 15/ 18
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
147/184
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region enclosed by the rose curve r= 3cos2.
Solution. The region is
We use symmetry.
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
148/184
e use sy et y.
Here, must vary from = 0 to =/4to sweep out the region. Thus
A = 8/4
0
1
2(3cos2)2 d= 36
/40
1
2(1 +cos4) d
= 18/40
(1 +cos4)d = 18+ 14
sin4/4
0
Polar Curves 15/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region enclosed by the rose curve r= 3cos2.
Solution. The region is
We use symmetry.
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
149/184
y y
Here, must vary from = 0 to =/4to sweep out the region. Thus
A = 8/4
0
1
2(3cos2)2 d= 36
/40
1
2(1 +cos4) d
= 18/40
(1 +cos4)d = 18+ 14
sin4/4
0= 9
2
Polar Curves 15/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region enclosed by the rose curve r= 3cos2.
Solution. The region is
We use symmetry.
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
150/184
y y
Here, must vary from = 0 to =/4to sweep out the region. Thus
A = 8/4
0
1
2(3cos2)2 d= 36
/40
1
2(1 +cos4) d
= 18/40
(1 +cos4)d = 18+ 14
sin4/4
0= 9
2
or = 4/4/4
1
2(3cos2)2 d
Polar Curves 15/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
151/184
Polar Curves 16/ 18
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
152/184
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.
Solution. The region is
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
153/184
Polar Curves 16/ 18
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
154/184
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.
Solution. The region is
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
155/184
We use symmetry.
Polar Curves 16/ 18
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
156/184
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.
Solution. The region is
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
157/184
We use symmetry. Here, must vary from = 0 to =.
Polar Curves 16/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.
Solution. The region is
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
158/184
We use symmetry. Here, must vary from = 0 to =.Solving for from the intersection of the two curves,
Polar Curves 16/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.
Solution. The region is
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
159/184
We use symmetry. Here, must vary from = 0 to =.Solving for from the intersection of the two curves,
4 +4cos= 6
Polar Curves 16/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.
Solution. The region is
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
160/184
We use symmetry. Here, must vary from = 0 to =.Solving for from the intersection of the two curves,
4 +4cos= 6 = cos = 12Polar Curves 16/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.
Solution. The region is
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
161/184
We use symmetry. Here, must vary from = 0 to =.Solving for from the intersection of the two curves,
4 +4cos= 6 = cos = 12 = = 3 , 53Polar Curves 16/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.
Solution. The region is
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
162/184
We use symmetry. Here, must vary from = 0 to =.Solving for from the intersection of the two curves,
4 +4cos= 6 = cos = 12 = = 3 , 53 = = 3Polar Curves 16/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.
Also, notice that the region can be obtained as follows.
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
163/184
Polar Curves 17/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.
Also, notice that the region can be obtained as follows.
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
164/184
=
Polar Curves 17/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.
Also, notice that the region can be obtained as follows.
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
165/184
=
Polar Curves 17/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.
Also, notice that the region can be obtained as follows.
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
166/184
= Hence, the area is
Polar Curves 17/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.
Also, notice that the region can be obtained as follows.
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
167/184
= Hence, the area is
A = 2/3
0
1
2(4+ 4cos)2 d
Polar Curves 17/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.
Also, notice that the region can be obtained as follows.
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
168/184
= Hence, the area is
A = 2/3
0
1
2(4+ 4cos)2 d 2
/30
1
2(6)2 d
Polar Curves 17/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.
Also, notice that the region can be obtained as follows.
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
169/184
= Hence, the area is
A = 2/3
0
1
2(4+ 4cos)2 d 2
/30
1
2(6)2 d =
/30
(16+ 32cos+ 16cos2 36)d
Polar Curves 17/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.
Also, notice that the region can be obtained as follows.
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
170/184
= Hence, the area is
A = 2/3
0
1
2(4+ 4cos)2 d 2
/30
1
2(6)2 d =
/30
(16+ 32cos+ 16cos2 36)d
= /3
0
(16cos2
+32cos
20)d
Polar Curves 17/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.
Also, notice that the region can be obtained as follows.
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
171/184
= Hence, the area is
A = 2/3
0
1
2(4+ 4cos)2 d 2
/30
1
2(6)2 d =
/30
(16+ 32cos+ 16cos2 36)d
= /3
0
(16cos2
+32cos
20)d
=18
3
4
Polar Curves 17/ 18
Tangent Lines to Polar Curves
Area in Polar Coordinates
Example.
Find the area of the region inside the cardioid r= 4 + 4cos and outside the circler= 6.
Also, notice that the region can be obtained as follows.
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
172/184
= Hence, the area is
A = 2/3
0
1
2(4+ 4cos)2 d 2
/30
1
2(6)2 d =
/30
(16+ 32cos+ 16cos2 36)d
= /3
0
(16cos2
+32cos
20)d
=18
3
4
or =/3/3
1
2(16cos2 + 32cos 20) d (without using symmetry)
Polar Curves 17/ 18
Tangent Lines to Polar Curves
Intersection of Polar Curves
Remark.
Intersection points of polar curves are naturally obtained by equating the
expressions for r.
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
173/184
Polar Curves 18/ 18
Tangent Lines to Polar Curves
Intersection of Polar Curves
Remark.
Intersection points of polar curves are naturally obtained by equating the
expressions for r. However, this method will not always produce all the
intersection points because of the many ways of representing a point in polarcoordinates.
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
174/184
Polar Curves 18/ 18
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
175/184
Tangent Lines to Polar Curves
Intersection of Polar Curves
Remark.
Intersection points of polar curves are naturally obtained by equating the
expressions for r. However, this method will not always produce all the
intersection points because of the many ways of representing a point in polarcoordinates.
Consider the cardioids r= 1 cos and r= 1 + cos. Equating the expressions,
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
176/184
q g p ,
1 cos= 1 +cos
Polar Curves 18/ 18
Tangent Lines to Polar Curves
Intersection of Polar Curves
Remark.
Intersection points of polar curves are naturally obtained by equating the
expressions for r. However, this method will not always produce all the
intersection points because of the many ways of representing a point in polar
coordinates.
Consider the cardioids r= 1 cos and r= 1 + cos. Equating the expressions,
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
177/184
q g p
1 cos= 1 +cos = cos= 0
Polar Curves 18/ 18
Tangent Lines to Polar Curves
Intersection of Polar Curves
Remark.
Intersection points of polar curves are naturally obtained by equating the
expressions for r. However, this method will not always produce all the
intersection points because of the many ways of representing a point in polar
coordinates.
Consider the cardioids r= 1 cos and r= 1 + cos. Equating the expressions,
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
178/184
g
1 cos= 1 +cos = cos= 0 = = 2
Polar Curves 18/ 18
Tangent Lines to Polar Curves
Intersection of Polar Curves
Remark.
Intersection points of polar curves are naturally obtained by equating the
expressions for r. However, this method will not always produce all the
intersection points because of the many ways of representing a point in polar
coordinates.
Consider the cardioids r= 1 cos and r= 1 + cos. Equating the expressions,
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
179/184
1 cos= 1 +cos = cos= 0 = = 2
+k2
, kZ.
Polar Curves 18/ 18
Tangent Lines to Polar Curves
Intersection of Polar Curves
Remark.
Intersection points of polar curves are naturally obtained by equating the
expressions for r. However, this method will not always produce all the
intersection points because of the many ways of representing a point in polar
coordinates.
Consider the cardioids r= 1 cos and r= 1 + cos. Equating the expressions,
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
180/184
1 cos= 1 +cos = cos= 0 = = 2
+k2
, kZ.
These values of give us only two points1, 2
and
1, 32
.
Polar Curves 18/ 18
Tangent Lines to Polar Curves
Intersection of Polar Curves
Remark.
Intersection points of polar curves are naturally obtained by equating the
expressions for r. However, this method will not always produce all the
intersection points because of the many ways of representing a point in polar
coordinates.
Consider the cardioids r= 1 cos and r= 1 + cos. Equating the expressions,
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
181/184
1 cos= 1 +cos = cos= 0 = = 2
+k2
, kZ.
These values of give us only two points1, 2
and
1, 32
.
Polar Curves 18/ 18
Tangent Lines to Polar Curves
Intersection of Polar Curves
Remark.
Intersection points of polar curves are naturally obtained by equating the
expressions for r. However, this method will not always produce all the
intersection points because of the many ways of representing a point in polar
coordinates.
Consider the cardioids r= 1 cos and r= 1 + cos. Equating the expressions,
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
182/184
1 cos= 1 +cos = cos= 0 = = 2
+k2
, kZ.
These values of give us only two points1, 2
and
1, 32
.
Clearly, the pole has been missed out.
Polar Curves 18/ 18
Tangent Lines to Polar Curves
Intersection of Polar Curves
Remark.
Intersection points of polar curves are naturally obtained by equating the
expressions for r. However, this method will not always produce all the
intersection points because of the many ways of representing a point in polar
coordinates.
Consider the cardioids r= 1 cos and r= 1 + cos. Equating the expressions,
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
183/184
1 cos= 1 +cos = cos= 0 = = 2
+k2
, kZ.
These values of give us only two points1, 2
and
1, 32
.
Clearly, the pole has been missed out. This is
because the cardioids pass through the pole
at different values of.
Polar Curves 18/ 18
Tangent Lines to Polar Curves
Intersection of Polar Curves
Remark.
Intersection points of polar curves are naturally obtained by equating the
expressions for r. However, this method will not always produce all the
intersection points because of the many ways of representing a point in polar
coordinates.
Consider the cardioids r= 1 cos and r= 1 + cos. Equating the expressions,
-
7/31/2019 Lecture 10 (Calculus of Polar Curves)
184/184
1 cos= 1 +cos = cos= 0 = = 2
+k2
, kZ.
These values of give us only two points1, 2
and
1, 32
.
Clearly, the pole has been missed out. This is
because the cardioids pass through the pole
at different values of.
In general, it is a good idea to graph thecurves to determine exactly how many
intersections points should there be.
Polar Curves 18/ 18